I am struggling with a portion of a proof concerning the lattice version of the Stone-Weierstrass theorem.
In particular, there is a subset $\mathcal{A}$ of the set of all real-valued continuous functions on a compact set $X$ such that if $f,g\in\mathcal{A}$ then $s\,f+g\in\mathcal{A}$ (where $s\in\mathbb{R}$) and $f\,g\in\mathcal{A.} $ Moreover, all constant functions are in $\mathcal{A}$.
I consider a function $g\in\mathcal{A}$ such that $\sup_{x\in X}|g(x)|\leq1$.
By an application of the Weierstrass approximation theorem, it holds true that for all $\delta>0$ there exists a polynomial $P(g)$ such that $$\sup_{g\in[-1,1]}\big||g|-P(g)\big|\leq\delta.$$
The proof now asserts that, since the constant functions are in $\mathcal{A}$, it follows that $P(\,g\,)\in\mathcal{A}.$
I am afraid, I do not see how this is true.
I would appreciate any help.
Many thanks.
Best Answer
As stated, the assertion is not true. Take $X=[-1,1]$ and $$ \mathcal A=\{\lambda 1 + f:\ f|_{[0,1]}=0\}. $$ Then $\mathcal A$ satisfies your conditions, and it contains no polynomials other than the constants.
Edit: with access to the link, it looks like you have misunderstood a few things. What happens is this:
the algebra $\mathcal A$ is closed;
lemma 5.10 guarantees that for any $\varepsilon>0$ there exists a polynomial $P_\varepsilon$ such that $$\tag{$*$}\sup_{s\in[-1,1]}|\,|s|-P_\varepsilon(s)|<\varepsilon.$$
as $|g|\leq1$, you have $g(t)\in[-1,1]$ for all $t$. Taking $s=g(t)$ in $(*)$ you get $$ |\,|g(t)|-P_\varepsilon(g(t))|<\varepsilon. $$ As the inequality holds for all $t\in X$, you get that $$\tag{$**$}\||\,|g|-P_\varepsilon(g)\|_\infty<\varepsilon. $$
Since $\mathcal A$ is an algebra that contains the constants you have, for any polynomial $p=\sum_{j=0}^ma_jx^j$, that $$ p(g)=a_0+a_1g+a_2g^2+\cdots+a_mg^m\in\mathcal A. $$